A motor car travelling to point pedro overtakes a cyclist at 9:00A.M.

IMO B.

Car Travels entire distance in 1.5 hours. so it speed, if the distance is x, is 2/3x
While returning, it meets the cyclist at noon i.e. 12 i.e. after driving for 1/2 hour.
In 30 mins car will cover 1/3x distance. So in 3 hours cyclist would have covered 2/3x distance.
2/3 ---> 3 hours
1 ---> 4.5 hours -- 1.30PM

Make a diagram to understand:

..................9:00........................................................................10:30
Car ------->------------------------------------------>
Cyclist-----------------------------><----------------<
...............................................................12 noon......................11:30


From the point at 9:00, the car reaches pedro at 10:30 i.e. in 1.5 hrs.
From pedro to point at 12 noon, the car reaches in 0.5 hrs. So pedro to this point is 1/3rd distance.
So the cyclist covered 2/3 rd distance in 3 hrs. He will cover 1/3rd distance in half the time i.e. 1.5 hrs. So he reaches pedro at 1:30.

My approach:

For motor car, distance from cyclist to Point Pedro be 3X, time taken=1.5 hours. Therefore Speed of motor car= 2X(units)/hour.

Now, the motor car meets the cyclist after half an hour. Hence distance covered by motor car= speed*time= 2X*0.5= X.
At this point, the cyclist has already travelled, 3X-X= 2X. Therefore speed of cyclist= 2X/3(units)/hour (note he travelled from 9 A.M. to 12 P.M.).

Remaining distance for cyclist to cover= X. Time=?. Speed= 2X/3. Therefore, distance= speed*time; Time=Distance/Speed =X/(2X/3) = 3/2= 1.5 hours. Hence 12 + 1.5 hours= 1.30 P.M.

Hence, B.

b - speed by cyclist, c - speed of card , X - refers to cyclist
|<---1.5b---X------------------->| - car (distance = 1.5c) at 10:30
|<--------------2.5b--X-----------| - car (distance = 1.5c) at 11:30

now both meet at noon, so \(\frac{{1.5c - 2.5b}}{{b + c}} = 1/2\) => C = 3b
time taken to travel 1.5c - 2.5b is 4.5b - 2.5b = 2b , 2b/b = 2 hours => 11:30 + 2 = 1:30

We can let the distance between the place (where the car overtakes the cyclist) and Point Pedro = d. Also, we can let the car’s speed = r and cyclist’s speed = s. Thus, we have:

1.5r = d

And

3s + 0.5r = d

Multiply the second equation by 3 we have:

9s + 1.5r = 3d

Subtract the first equation from this, we have

9s = 2d

4.5s = d

Thus it takes 4.5 hours for the cyclist to reach Point Pedro from the place (where he is overtaken by the car). Thus he reaches Point Pedro at 9:00 AM + 4.5 hours = 13:30 = 1:30 PM.

Answer: B

The Car takes 1.5 hours to get from the Point that the Cyclist is at to Point Pedro (PP).

The Distance from where the Cyclist is to PP = 1.5 hours (R) ----- where R is the Speed of the Car


When the Car Turns around to meet the Cyclist, he only travels a Distance = .5 hours (R)


This means the Distance covered by the Bicyclist from where she was Passed by the Motorist to the Meet Point at 12 PM = 1.5R Total Distance measured by Car's Speed - .5R that the Car has traveled to meet the Cyclist = 1R

As a Fraction of the Total Distance, from 9 AM to 12 PM (3 hours) the Cyclist Traveled a Distance of = 1R / 1.5R = 1 / 1.5 = 10 / 15 = 2/3rd of Distance


Cyclist has 1/3rd of the Distance left to travel until he reaches PP. Cyclist traveled 2/3rd of Distance in 3 Hours (1.5 hours car took to PP, + 1 hour wait + 1/2 hour to meet Cyclist at Noon)

Speed of Cyclist = 2/3 of Total Distance / 3 hours = (2/3) / 3 = 2/9 of the Total Distance per hour

Distance Left to Travel for Cyclist = 1/3 of Distance

Time = Distance / Speed

Time = (1/3) / (2/9) = 9 / 6 = 1.5 hours until the Cyclist reaches PP after he meets the Car at NOON

12 PM + 1.5 hours = 1:30 PM

The Cyclist will arrive at 1:30 PM at PP

-Answer B-

If we take distance as 1 and total time taken by motor car to reach pedro is 1hr 30 mins that is 90 mins thus it speed = 1/90

Motor car driver after waiting for an hour , he left at 11:30 and meet again with cyclist at 12:00 thus means , motor car driver travel for 30 mins before meeting again with cyclist.  

Thus distance covered by Motor car driver in 30 mins = 1/90 * 30 = 1/3

Thus it means cyclict already covered 1-1/3 = 2/3 distance in 3 hours (180 mins)

so it means cyclist speed = (2/3) / 180 = 1/270

now cyclist need to cover more 1/3 distance to reach pedro thus cyclist need to travel for Time = (1/3) / (1/270) = 270/3 = 90 mins

Total Time Cyclist took to reach pedro = 180+90 = 270 mins that is 4hrs 30 min 
Cyclist started at 9 :00 AM thus after 4hr 30 min the time will be 1:30 PM thus B

The question does not say the speed is constant or it is what we have to assume ourselves?

­
Without making that assumption, the question would be unsolvable. Nevertheless, I agree that the question could have been worded better.

Speed of motor car m = \(\frac{d}{90}\) (distance "d" by 90 minutes)

and distance covered by car in 30min return journey = m*30mins or you can say \(\frac{d}{3}\) (since, we just established 90*m =d => 30*m = \(\frac{d}{3}\))

Now, distance travelled by cyclist to meet the motor car on its return
=> speed of cyclist c * (90 + 60 + 30) (90 minutes travel time by motorcar + 60 min wait + 30 min return time by motorcar)
=> 180*c + \(\frac{d}{3}\) = total distance d (cyclist journey + motorcar's return journey distances = total journey)
=> 180*c = d - \(\frac{d}{3}\) = \(\frac{2*d}{3}\)

=> c = \(\frac{d}{270}\) => means cyclist covers d distance in 270 minutes

=> 270 minutes = 4 hours + 30mins (9:00am + 4:30mins = 1:30pm)

B. 1:30PM